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6.1 Q-8

Question Statement

We are given two circles with the following equations:

  1. x2+y2+2xβˆ’8=0x^2 + y^2 + 2x - 8 = 0
  2. x2+y2βˆ’6x+6yβˆ’46=0x^2 + y^2 - 6x + 6y - 46 = 0

We need to show that these two circles touch internally.


Background and Explanation

To determine if two circles touch internally, we need to check the following condition:

  • The distance between their centers should be equal to the difference in their radii.

Key Concepts:

  1. The center of a circle in the form x2+y2+2gx+2fy+c=0x^2 + y^2 + 2gx + 2fy + c = 0 is (βˆ’g,βˆ’f)(-g, -f).
  2. The radius of a circle is given by: r=g2+f2βˆ’cr = \sqrt{g^2 + f^2 - c}

We will use these formulas to find the centers and radii of both circles, then check if the distance between their centers equals the difference in their radii.


Solution

Step 1: Extract the centers and radii of both circles

Circle (1): x2+y2+2xβˆ’8=0x^2 + y^2 + 2x - 8 = 0

  • The equation is in the general form x2+y2+2gx+2fy+c=0x^2 + y^2 + 2gx + 2fy + c = 0, where:
    • g=1g = 1, f=0f = 0, and c=βˆ’8c = -8.
  • The center of the circle is (βˆ’g,βˆ’f)=(βˆ’1,0)(-g, -f) = (-1, 0).
  • The radius is:
r1=g2+f2βˆ’c=12+02βˆ’(βˆ’8)=1+8=3 r_1 = \sqrt{g^2 + f^2 - c} = \sqrt{1^2 + 0^2 - (-8)} = \sqrt{1 + 8} = 3

Circle (2): x2+y2βˆ’6x+6yβˆ’46=0x^2 + y^2 - 6x + 6y - 46 = 0

  • The equation is in the form x2+y2+2gx+2fy+c=0x^2 + y^2 + 2gx + 2fy + c = 0, where:
    • g=βˆ’3g = -3, f=3f = 3, and c=βˆ’46c = -46.
  • The center of the circle is (βˆ’g,βˆ’f)=(3,βˆ’3)(-g, -f) = (3, -3).
  • The radius is:
r2=g2+f2βˆ’c=(βˆ’3)2+32βˆ’(βˆ’46)=9+9+46=64=8 r_2 = \sqrt{g^2 + f^2 - c} = \sqrt{(-3)^2 + 3^2 - (-46)} = \sqrt{9 + 9 + 46} = \sqrt{64} = 8

Step 2: Calculate the distance between the centers

The distance dd between the centers (βˆ’1,0)(-1, 0) and (3,βˆ’3)(3, -3) is given by the distance formula:

d=(x2βˆ’x1)2+(y2βˆ’y1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Substitute the coordinates of the centers:

d=(3βˆ’(βˆ’1))2+(βˆ’3βˆ’0)2=(3+1)2+(βˆ’3)2=42+(βˆ’3)2=16+9=25=5d = \sqrt{(3 - (-1))^2 + (-3 - 0)^2} = \sqrt{(3 + 1)^2 + (-3)^2} = \sqrt{4^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5

Step 3: Check the condition for internal tangency

For the circles to touch internally, the distance between the centers should be equal to the difference in their radii:

r2βˆ’r1=8βˆ’3=5r_2 - r_1 = 8 - 3 = 5

Since the distance between the centers is 5, which is equal to the difference in the radii, we can conclude that the circles touch internally.


Key Formulas or Methods Used

  • Center of a Circle:
    From the general equation x2+y2+2gx+2fy+c=0x^2 + y^2 + 2gx + 2fy + c = 0, the center is (βˆ’g,βˆ’f)(-g, -f).

  • Radius of a Circle:
    The radius is given by: r=g2+f2βˆ’cr = \sqrt{g^2 + f^2 - c}

  • Distance Between Two Points:
    The distance between the centers (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) is: d=(x2βˆ’x1)2+(y2βˆ’y1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}


Summary of Steps

  1. Step 1: Find the centers and radii of both circles.
  2. Step 2: Calculate the distance between the centers of the two circles.
  3. Step 3: Check if the difference in the radii is equal to the distance between the centers.
  4. Step 4: Conclude that the circles touch internally because the difference in the radii equals the distance between their centers.